Real Numbers 🔢

Euclid ka algorithm, prime factorisation, HCF, LCM, aur irrational numbers ke saare proofs.

🌐 Number System — Ek Badi Tasveer

🤔 Socho aise:

Counting karte waqt hum 1, 2, 3... use karte hain — ye Natural Numbers hain. Jab 0 add kiya — Whole Numbers. Jab negative bhi aaye — Integers. Jab fraction aaye — Rational Numbers. Aur jo fraction mein express hi nahi ho sakti — woh Irrational Numbers. Sab mila ke — Real Numbers!

Natural: 1,2,3,4... Whole: 0,1,2,3... Integer: ...−2,−1,0,1,2... Rational: p/q form Irrational: √2, π, e Real = Rational ∪ Irrational
→ −3 −2 −1 0 1 2 3 √2≈1.41 π≈3.14

1️⃣ Euclid ka Division Lemma (Vibhaajan Pramaeyika)

Theorem: Koi bhi do positive integers a aur b diye jaayein, to unique integers q (quotient) aur r (remainder) exist karte hain jaise:

a = bq + r ,   0 ≤ r < b
📦 Real-life example:

Tumhare paas 47 chocolates hain aur 5 friends ko equally baantni hain. 47 ÷ 5 = 9 remainder 2. Matlab: 47 = 5×9 + 2. Yahi Euclid ka Division Lemma hai!

Example: a=135, b=12
135 = 12 × 11 + 3
Here q=11, r=3. Check: 12×11+3 = 132+3 = 135 ✓

2️⃣ Euclid ka Division Algorithm — HCF Nikalna

Euclid ka lemma use karke hum do numbers ka HCF (Highest Common Factor) nikal sakte hain:

Algorithm
Step 1: a = bq + r likhो (a > b)
Step 2: Agar r = 0, to HCF = b
Step 3: Agar r ≠ 0, to naya problem: HCF(b, r) karo — wapas Step 1
Example: HCF(455, 42) nikalo
Step 1
455 = 42 × 10 + 35  (r=35 ≠ 0)
Step 2
42 = 35 × 1 + 7  (r=7 ≠ 0)
Step 3
35 = 7 × 5 + 0  (r=0 ✓)
∴ HCF(455, 42) = 7
Example: HCF(867, 255)
867 = 255×3 + 102
255 = 102×2 + 51
102 = 51×2 + 0
HCF = 51

3️⃣ Fundamental Theorem of Arithmetic (FTA)

FTA: Har composite number ko unique way mein prime numbers ka product likha ja sakta hai (order ki parwah kiye bina).

Ye ek bahut powerful theorem hai — isi ki wajah se HCF aur LCM nikalna possible hai!

Factor Tree: 360 = ?

360 2 180 2 90 2 45 9 5 360 = 2³ × 3² × 5
HCF aur LCM — Prime Factorisation Method:
HCF = sabse chhotey powers ka product (common primes mein)
LCM = sabse bade powers ka product (sab primes mein)
HCF × LCM = a × b (do numbers ke liye)
Example: HCF aur LCM of 12 and 18
12 = 2² × 3
18 = 2 × 3²

HCF = 2¹ × 3¹ = 6 (minimum powers)
LCM = 2² × 3² = 36 (maximum powers)
Check: HCF × LCM = 6 × 36 = 216 = 12 × 18 ✓
Example: HCF aur LCM of 96 aur 404
96 = 2⁵ × 3
404 = 2² × 101
HCF = 2² = 4
LCM = 2⁵ × 3 × 101 = 9696
Check: 4 × 9696 = 38784 = 96 × 404 ✓

4️⃣ Irrational Numbers ke Proofs

Class 9 mein tumne √2 ka irrational proof seekha tha. Ab Class 10 mein hum zyada numbers prove karte hain ki woh irrational hain — aur contradiction method use karte hain.

Contradiction Method (Kya hai?): Hum pehle assume karte hain ki number rational hai (yaani p/q form mein, jahan p aur q coprime integers hain, q≠0). Phir dikhate hain ki yeh assumption galat hai — contradiction aata hai. Therefore number irrational hai.

Proof 1: √2 Irrational Hai

Assume
√2 rational hai. To √2 = p/q jahan HCF(p,q)=1
Square karo
2 = p²/q² → p² = 2q²
Conclusion
p² even hai → p even hai (kyunki odd ka square odd hota hai)
p=2m likhein
(2m)² = 2q² → 4m² = 2q² → q² = 2m²
Contradiction
q² even → q even. Par ab p aur q dono even hain — HCF(p,q)≥2. Yeh humari assumption ke viruddh hai!
Conclusion
∴ √2 irrational hai ■

Proof 2: √3 Irrational Hai

Assume √3 = p/q (HCF=1). Square karo: p²=3q² → p divisible by 3 → p=3m → 9m²=3q² → q²=3m² → q divisible by 3. Ab p aur q dono 3 se divisible — contradiction! ∴ √3 irrational

Proof 3: √5 Irrational Hai (same method)

Assume √5 = p/q. p²=5q² → p=5m → q²=5m² → q divisible by 5. Contradiction! ∴ √5 irrational

Proof 4: 3 + 2√5 Irrational Hai (Given √5 irrational)

Assume
3 + 2√5 rational hai. To 3 + 2√5 = r (rational number)
Rearrange
2√5 = r − 3 → √5 = (r−3)/2
Contradiction
RHS rational hai (rational numbers ka arithmetic rational hota hai). Par √5 irrational hai — contradiction!
Conclusion
∴ 3 + 2√5 irrational hai ■

5️⃣ Rational Numbers ka Decimal Expansion

Theorem: p/q (simplest form) ka decimal expansion terminating hoga agar q = 2ⁿ × 5ᵐ (sirf 2 aur 5 ke powers) — warna non-terminating repeating.
Numberq ki factorisationDecimal TypeValue
7/88 = 2³Terminating ✓0.875
1/66 = 2×3Non-terminating repeating0.1666...
13/125125 = 5³Terminating ✓0.104
17/66 = 2×3Non-terminating repeating2.8333...
7/2020 = 2²×5Terminating ✓0.35
Yaad rakhne ki baat: Irrational numbers ka decimal expansion non-terminating aur non-repeating hota hai — jaise π = 3.14159265..., √2 = 1.41421356...

6️⃣ Practice Questions

Q1. Euclid ke algorithm se HCF(336, 54) nikalo.

336 = 54 × 6 + 12
54 = 12 × 4 + 6
12 = 6 × 2 + 0
HCF = 6

Q2. 26 aur 91 ka HCF aur LCM prime factorisation se nikalo. Verify karo ki HCF×LCM = a×b.

26 = 2 × 13;   91 = 7 × 13
HCF = 13,   LCM = 2 × 7 × 13 = 182
Check: 13 × 182 = 2366 = 26 × 91 ✓

Q3. 17/8 aur 64/455 — in mein se kaun terminating decimal hai?

8 = 2³ → only factor 2 → 17/8 terminating = 2.125
455 = 5 × 7 × 13 → has 7 and 13 → 64/455 non-terminating repeating

Q4. Prove karo ki 5 − 2√3 irrational hai.

Assume 5−2√3 = r (rational)
2√3 = 5−r → √3 = (5−r)/2
RHS rational, par √3 irrational — contradiction!
∴ 5−2√3 irrational hai ■

Q5. Ek dakiya 18 minute mein aur doosra 24 minute mein ek chakkar lagaata hai. Dono ek saath start karte hain — kitne minute baad phir milenge?

LCM(18, 24): 18=2×3², 24=2³×3. LCM = 2³×3² = 72 minutes
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